Ask question

Ask question

All categories

3 years ago

6

Which is a chemical reaction? Select one: a. a peanut shell cracking open into two halves b. solid chocolate melting into liquid

chocolate c. water vapor condensing into liquid water on a window d. hydrogen gas and oxygen gas synthesizing into liquid water

2 answers:

raketka [301]3 years ago

8 0

My best educated guess would be d

sveta [45]3 years ago

7 0

I agree ^^ I think it would be d

You might be interested in

A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$

3 0

3 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

A small solid sphere and a small thin hoop are rolling along a horizontal surface with the same translational speed when they en

torisob [31]

This situation has a basis such that the solid sphere and the hoop has the same mass. The analysis could be made<span> backwards . The ball will decelerate fastest, so not roll as high. The sphere will accelerate faster, but this also means it decelerates faster on the way up. Hence the answer is the hoop if the masses are equal </span>

3 0

3 years ago

The velocity of a mass moving in a circular path is..

Alenkasestr [34]

Constantly changing.

5 0

3 years ago

Science and Technology are interdependent. Advances in one lead to advances in the other. Give an example of this phenomenon

Pepsi [2]

If more microscopic technology is invented, then more things in the world can be discovered. Likewise, if a new element is discovered, then people can use it in their technology.

7 0

3 years ago